If we are given a trigonometric expression as $\cos \left( \theta +\varnothing \right)=m\cos \left( \theta -\varnothing \right)$ then $\tan \theta $ is equal to
\[\begin{align}
  & A.\left( \dfrac{1+m}{1-m} \right)\tan \varnothing \\
 & B.\left( \dfrac{1-m}{1+m} \right)\tan \varnothing \\
 & C.\left( \dfrac{1+m}{1-m} \right)\cot \varnothing \\
 & D.\left( \dfrac{1-m}{1+m} \right)\cot \varnothing \\
\end{align}\]

Hint: In this question, we are given a trigonometric expression and we have to find the value of $\tan \theta $ in terms of m and angle $\varnothing $. For simplifying this we will apply componendo dividendo. After that, we will use the trigonometric formula of $\cos C+\cos D$ and $\cos C-\cos D$ to simplify calculation. At last, we will apply the following trigonometric formulas to obtain our answer.
\[\dfrac{\cos \theta }{\sin \theta }=\cot \theta \text{ and }\dfrac{1}{\tan \theta }=\cot \theta \]
Formula of $\cos C+\cos D$ is given as,
\[\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\]
Formula of $\cos C-\cos D$ is given as,
\[\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)\]

Complete step-by-step solution:
Here, we are given the equation as, $\cos \left( \theta +\varnothing \right)=m\cos \left( \theta -\varnothing \right)$ and we have to find the value of $\tan \theta $. For this, let us simplify the given equation. Firstly, let us bring all cosine terms on one side and 'm' terms on other, we get:
\[\dfrac{\cos \left( \theta +\varnothing \right)}{\cos \left( \theta -\varnothing \right)}=m\]
Let us now apply componendo and dividendo to above equation, we get:
\[\Rightarrow \dfrac{\cos \left( \theta +\varnothing \right)+\cos \left( \theta -\varnothing \right)}{\cos \left( \theta +\varnothing \right)-\cos \left( \theta -\varnothing \right)}=\dfrac{m+1}{m-1}\]
Let us take negative sign common from denominator of both sides, we get:
\[\Rightarrow \dfrac{\cos \left( \theta +\varnothing \right)+\cos \left( \theta -\varnothing \right)}{-\left( \cos \left( \theta -\varnothing \right)-\cos \left( \theta +\varnothing \right) \right)}=\dfrac{m+1}{-\left( 1-m \right)}\]
Cancelling the negative sign from denominator of both sides and rearranging numerator of right hand side, we get:
\[\Rightarrow \dfrac{\cos \left( \theta +\varnothing \right)+\cos \left( \theta -\varnothing \right)}{\cos \left( \theta -\varnothing \right)-\cos \left( \theta +\varnothing \right)}=\dfrac{1+m}{1-m}\]
As we know, $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\text{ and }\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$ so applying them on left side, we get:
\[\Rightarrow \dfrac{2\cos \left( \dfrac{\theta +\varnothing +\theta -\varnothing }{2} \right)\cos \left( \dfrac{\theta +\varnothing -\theta +\varnothing }{2} \right)}{2\sin \left( \dfrac{\theta +\varnothing +\theta -\varnothing }{2} \right)\sin \left( \dfrac{\theta +\varnothing -\theta +\varnothing }{2} \right)}=\dfrac{1+m}{1-m}\]
Simplifying the angles of cosine and sine, we get:
\[\Rightarrow \dfrac{2\cos \theta \cos \varnothing }{2\sin \theta \sin \varnothing }=\dfrac{1+m}{1-m}\]
Cancelling 2 from numerator and denominator of left side, we get:
\[\Rightarrow \dfrac{\cos \theta \cos \varnothing }{\sin \theta \sin \varnothing }=\dfrac{1+m}{1-m}\]
As we know, $\dfrac{\cos \theta }{\sin \theta }=\cot \theta $ and similarly $\dfrac{\cos \varnothing }{\sin \varnothing }=\cot \varnothing $. Therefore, applying it to left side we get:
\[\Rightarrow \cot \theta \cot \varnothing =\dfrac{1+m}{1-m}\]
Applying $\dfrac{1}{\tan \theta }=\cot \theta $ on left side we get:
\[\Rightarrow \dfrac{\cot \varnothing }{\tan \theta }=\dfrac{1+m}{1-m}\]
Taking reciprocal on both sides we get:
\[\Rightarrow \dfrac{\tan \theta }{\cot \varnothing }=\dfrac{1-m}{1+m}\]
Now, we have to find value of $\tan \theta $ so taking $\cot \varnothing $ on other side we get:
\[\Rightarrow \tan \theta =\left( \dfrac{1-m}{1+m} \right)\cot \varnothing \]
Hence, option D is the correct answer.

Note: Students should note that, while applying componendo-dividendo we have taken 1 as the denominator of m and hence, added m and 1 in the numerator and subtract them in the denominator. Students should keep in mind all the trigonometric formulas and identities. Apply $\cos C+\cos D$ and $\cos C-\cos D$ formula carefully and don't make mistakes in the plus-minus sign.